If you have 39 g of sodium azide in an air bag [2nan3 (s) - > 2na (s) + 2n2 (g) ], then how many molecules of nitrogen gas will it make at stp?

Sodium Azide in an air bag = 39g

Converting the given grams in to moles, that is 1 mole = 65 grams

Sodium Azide in moles = 0.6

From the equation there are 2 moles of Sodium Azide and there are 3 moles of N

Calculating the N2 in moles = (3 x 0.6) / 2 = 0.9 moles of N2

Calculating the molecular mass of N = 0.90 x 6.022 x 10^23 = 5.42 x 10^23.

2nan3 - - >2na + 3h2

calculate the moles of nan3 = 39/65=0.6 moles

the reacting ratio of nan3 to h2 is 2:3 therefore the mole of h2 formed is (0.6 x3) / 2=0.9 moles

using the avogadros law

1 mole=6.02 x 10^23, what about 0.9 moles

=0.9 x (6.02 x10^23) = 5.418 x 10^23